float modff( float arg, float* iptr );
(1) (since C99)double modf( double arg, double* iptr );
(2)long double modfl( long double arg, long double* iptr );
(3) (since C99)1-3) Decomposes given floating-point value arg into integral and fractional parts, each having the same type and sign as arg. The integral part (in floating-point format) is stored in the object pointed to by iptr.
[edit] Parameters arg - floating-point value iptr - pointer to floating-point value to store the integral part to [edit] Return valueIf no errors occur, returns the fractional part of arg with the same sign as arg. The integral part is put into the value pointed to by iptr.
The sum of the returned value and the value stored in *iptr gives arg (allowing for rounding).
[edit] Error handlingThis function is not subject to any errors specified in math_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
This function behaves as if implemented as follows:
[edit] Example#include <float.h>
#include <math.h>
#include <stdio.h>
int main(void)
{
double f = 123.45;
printf("Given the number %.2f or %a in hex,\n", f, f);
double f3;
double f2 = modf(f, &f3);
printf("modf() makes %.2f + %.2f\n", f3, f2);
int i;
f2 = frexp(f, &i);
printf("frexp() makes %f * 2^%d\n", f2, i);
i = ilogb(f);
printf("logb()/ilogb() make %f * %d^%d\n", f / scalbn(1.0, i), FLT_RADIX, i);
// special values
f2 = modf(-0.0, &f3);
printf("modf(-0) makes %.2f + %.2f\n", f3, f2);
f2 = modf(-INFINITY, &f3);
printf("modf(-Inf) makes %.2f + %.2f\n", f3, f2);
}
Possible output:
Given the number 123.45 or 0x1.edccccccccccdp+6 in hex, modf() makes 123.00 + 0.45 frexp() makes 0.964453 * 2^7 logb()/ilogb() make 1.92891 * 2^6 modf(-0) makes -0.00 + -0.00 modf(-Inf) makes -INF + -0.00[edit] References
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